Question: A particle moves along the curve $\dfrac xy=-9$ so that that the $y$ -coordinate is increasing at a constant rate of $1$ unit per minute. What is the magnitude (in units per minute) of the particle's velocity vector when the particle is at the point $(9,-1)$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $2$ (Choice B) B $9$ (Choice C) C $\sqrt{82}$ (Choice D) D $4\sqrt{5}$
Background When working with motion along a curve, we should remember that this motion can also be represented by a position vector $(x,y)$. The main difference is that we're not given the equations for $x$ and $y$ in terms of time $t$, but only the relationship between $x$ and $y$ themselves. This however shouldn't prevent us from assuming the expressions for $x$ and $y$ in terms of $t$ exist. As always, $\vec{v}(t)=\left(\dfrac{dx}{dt},\dfrac{dy}{dt}\right)$. Setting up the math We are given that $\dfrac{dy}{dt}=1$ for any value of $t$. We are asked for the magnitude of the particle's velocity vector when the particle is at the point $(9,-1)$. In other words, we are asked for $\left|\left|\left(\dfrac{dx}{dt},\dfrac{dy}{dt}\right)\right|\right|$ at the point $(9,-1)$. Finding $\dfrac{dx}{dt}$ $\dfrac{dx}{dt}=\dfrac{x}{y}$ Finding $\dfrac{dx}{dt}$ at $(9,-1)$ The expression for $\dfrac{dx}{dt}$ depends on both the particle's $x$ -coordinate ${9}$ and its $y$ -coordinate ${-1}$ : $\begin{aligned} \dfrac{dx}{dt}&=\dfrac{({9})}{({-1})} \\\\ &=-9 \end{aligned}$ Therefore, the particle's velocity vector at the point $(9,-1)$ is $(-9,1)$. Finding $||(-9,1)||$ $||(-9,1)||=\sqrt{82}$ In conclusion, the magnitude of the particle's velocity vector when the particle is at the point $(9,-1)$ is $\sqrt{82}$ units per minute.